# Finding the area of 4 rectangles

I am stuck on another optimization problem and I can't get my answer to match the author's. I am assuming the author is correct, but there is no justification for their answer.

A farmer with 750 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens?

I know that the two formulas I need are

$2w + 5l = 750$ w for width and l for length to find the perimeter. I know it is this because there are 3 fences laid down inside a rectangle, so that gives me 5 lines of fence in one way and then 2 in the other direction.

$4lw = a$ for this I know that I have 4 rectangles so it is 4 lw.

Subsituting in w in terms of l

$2l(750-5l) = a$

$1500l-10l^2=a$

then I take the derivative

$1500 - 20l = a$

$l=75$ which gives me the wrong answer.

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"then I take the derivative--$1500−20l=a$" --Without going into the details of whether the previous steps are right, this step is wrong, for a constant $a$. it's derivative is $0$. But, I am surprised you conclude rightly! – user21436 Apr 4 '12 at 0:00
You are using $l$ and $w$ to mean the length and width of the larger, outermost rectangle, not the length and width of the smaller rectangles resulting from the subdivision. This means the total area would be just $lw$, not $4lw$. – Austin Mohr Apr 4 '12 at 0:00
@Austin I don't understand, I set up the problem to be 5w and 2l so the cooresponding area would use the same shapes. I suppose it would be 1/4w and 4l though. – user138246 Apr 4 '12 at 0:04

Your mistake seems to be that you're confusing the width of each pen with the width of the whole area. The formula $2w + 5l = 750$ assumes that $w$ is the width of all four pens added together, yet $4lw = a$ treats $w$ as the width of only one pen. Once you make the definition of $w$ consistent, you should get the right answer.

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Always draw a picture. The pens should look something like this:

Total length of fencing: $5l+2w=750 \mbox{ ft}$

$Area=l\times w =\frac{750}{2}l-\frac{5}{2}l^2\mbox{ ft}^2$

To find the value of $l$ that maximises the area we differentiate the area formula above with respect to $l$ and set it to zero, then solve for $l$

$\frac{dArea}{dl}=\frac{750}{2}-5l=0$

which has solution $l=75 \mbox{ft}$ and so the required maximum area is $$Area_{max}=\frac{750}{2}75-\frac{5}{2}75^2=14062.5\mbox{ ft}^2$$ (that this is a maximum you can confirm by taking the second derivative of the area with respect to $l$ and examining its sign)

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